Consider the coordinate plane. Let $A=(0,0)$, $B=(1,0)$, $C=(0,1)$, $D=(1,1)$ and $X=(u,v)$. Then: $$\sqrt{u^2+v^2}=AX,\quad \sqrt{(u-1)^2+v^2}=BX,\quad \sqrt{u^2+(v-1)^2}=CX,\quad \sqrt{(u-1)^2+(v-1)^2}=XD$$Using the triangle inequality for (the possibly degenerate triangles) $\triangle AXD$ and $\triangle BXC$ we have that: $$\sqrt{u^2+v^2} +\sqrt{(u-1)^2+v^2}+\sqrt {u^2+ (v-1)^2}+ \sqrt{(u-1)^2+(v-1)^2}=AX+BX+CX+DX\geq AD+BC=2\sqrt{2}=\sqrt{8}$$Equality occurs iff $X\in AD$ and $X\in BC$, so only if $(u,v)=\left(\frac{1}{2},\frac{1}{2}\right)$ and for these values of $u,v$ the expression does equal the obtained minimum. Therefore $n=8$ and the answer is $\boxed{80}$.

By Minkowski's Inequality, $$\sqrt{u^2+v^2} +\sqrt{(1-u)^2+v^2}+\sqrt {u^2+ (1-v)^2}+ \sqrt{(1-u)^2+(1-v)^2}\geq\sqrt{(u+(1-u)+u+(1-u))^2+(v+v+(1-v)+(1-v))^2}= \sqrt{2^2+2^2}=\sqrt{8}$$with $(u,v)=\left(\frac{1}{2},\frac{1}{2}\right)$ as an equality case. Thus, $10n=10\cdot 8=80$.

By QM-AM, $$\sqrt{u^2+v^2} +\sqrt{(u-1)^2+v^2}+\sqrt {u^2+ (v-1)^2}+ \sqrt{(u-1)^2+(v-1)^2} \geq \sqrt{2}(|u|+|v|+|v-1|+|u-1|).$$It's easy to show that the minimum of this is $2\sqrt 2=\sqrt 8$. So the answer is $80$.